Assuming that $V$ is a vector space, and $U$ is assumed to be a subspace of $V$, your proof has the right gist, but it's a bit sloppy. The first thing should be something like:
(reflexive) We need to show that for all $v \in V$, there's a $u \in U$ with $v+u = v$. Because $U$ is a subspace, and therefore contains the 0-vector, we can choose $u = 0$ and get $v + u = v + 0 = v$.
I grant you that's a bit wordy, but it does state explicitly what needs to be shown, and then explains why $0$ is allowed as a possible "u" in the proof.
For the third one, I'll get you started:
(transitive) Suppose that $v \sim w$ and $w \sim x$; we need to show that $v \sim x$, i.e., that there's a vector $u \in U$ with $v + u = x$. From the first assumption, we know there's a vector $u_1 \in U$ with $v + u_1 = w$. From the second...
Of course, both of these are proofs suitable for someone who's just learning linear algebra. In a research paper, it would probably suffice to say something like "Because $U$ is a subspace of $V$, the relation $v \sim w$ iff $v-w \in U$ is evidently an equivalence relation." So there is (as usual in mathematics) a question of audience for any given proof.